This is my code:
set-location [PATH]
$A = Start-Process -FilePath .\refresh.bat -Wait
set-location C:\
When executed in powershell, the system opens a Command prompt window and executes the bat file without issue. The problem is that the window closes and I cannot see if there was an error if it succeeds.
I want to keep the CMD window open.
I also tried at the end of the bat file:
:END
cmd /k
but no luck.
First, unless you specifically need to run the batch file in a new window, do not use Start-Process - use direct invocation instead, which is implicitly synchronous and allows you to capture or redirect output:
# Invoke the batch file synchronously (wait for it to exit)
# and capture its (standard) output in variable $A
# To print the batch file's output to the console instead, just use:
# .\refresh.bat
$A = .\refresh.bat
See this answer for more information.
Also note Start-Process never allows you to capture the invoked program's output directly (you can only redirect it to files with -RedirectStandardOutput and -RedirectStandardOutput); your specific command captures nothing[1] in $A; adding -PassThru does return something, but not the program's output, but a process-information object (System.Diagnostics.Process).
If you do need to run the batch file in a new window and want to keep that window open:
Start-Process -Wait -FilePath cmd -ArgumentList '/k .\refresh.bat'
Relying on positional parameter binding, the above can be simplified to:
Start-Process -Wait cmd '/k .\refresh.bat'
[1] Strictly speaking, $A is assigned the [System.Management.Automation.Internal.AutomationNull]::Value singleton, which in most contexts behaves like $null.
Thank you mklement0 with your post gave me the solution I wanted. This is how I solved it.
set-location [PATH]
$A = Start-Process -FilePath .\refresh.bat -Wait -NoNewWindow
set-location C:\
-NoNewWindow allowed me to run my batch in the same powershell window getting the feedback of the bat file. That way I have errors if any and success status if no errors.
Thanks!